## Sunday, July 8, 2012

### Problem 779: Triangle, Altitude, Orthocenter, Vertex, Midpoint, Side, Angle, 90 Degrees

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 779.

1. By properties of orthocenter, and cyclic quadrilateral,
Angle DAC = Angle DHB

Thus,
Triangle DAC ~ Triangle DHB

Now, consider a rotation 90 degrees clockwise about point D,
DB maps to a line segment on DC,
DH maps to a line segment on DA.

So, after the above rotation,
triangle DHB will become the same position as triangle DAC, only with different size.

Hence, DM will map to a line segment on DF.
Which means, angle FDM = 90 degrees.

2. Triangle ADC is similar to triangle DHB
And since F is the mid point of AC and M is the mid point of BH, so triangle ADF is similar to DHM and DFC is similar to DHM
So angle FDM is same as angle ADC , which is 90
QED

3. http://img37.imageshack.us/img37/6047/problem779.png

Draw nine-points circle per attached sketch
This circle will pass through feet of altitudes, midpoint of each side and midpoint from orthocenter to each vertex.
Since ∠ (MNF)=90 => MF is a diameter
So ∠ (MDF)=90

4. ∆ADC is rt angled at D, and F is the midpoint of the hypotenuse AC.
So AF = DF, ∠ADF = A.
∆BDH is rt angled at D, and M is the midpoint of the hypotenuse BH.
So ∠BDM = ∠DBM = 90° - A (note BH ⊥ AC)
Follows ∠ADF + ∠BDM = 90°
Hence ∠FDM = 90°